Chapter 11, Problem 36RP

a.

To determine

To calculate:Thearea of shaded region with arc radius as 10.

Answer to Problem 36RP

Thearea of shaded region is $16.12$ .

Explanation of Solution

Given information:

Radius of arc = 10

Formula used:

Area of equilateral triangle: $A=\frac{s^2}{4}\sqrt{3}$

s = side of equilateral triangle.

Area of sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$

r = radius of circle.

Calculation:

  

Draw an equilateral triangle, and from the area of the largest triangle, subtract the areas of the three sectors.

  $\begin{array}{l}{A}_{Triangle}=\frac{s^2}{4}\sqrt{3}\\ A_{Triangle}=\frac{{20}^2}{4}\sqrt{3}\\ A_{Triangle}=100\sqrt{3}\end{array}$

  $\begin{array}{l}{A}_{sector}=\left(\frac{measureofarc}{360^o}\right)\pi r^2\\ A_{sector}=\left(\frac{{60}^{\circ }}{360^o}\right)\pi 10^2\\ A_{sector}=\left(\frac{1}{6}\right)100\pi \\ A_{sector}=\frac{50}{3}\pi \end{array}$

There are 3 sectors.

Area shaded $=A_{Triangle}-3A_{\mathrm{Sec}tor}$

  $\begin{array}{l}=100\sqrt{3}-3\left(\frac{50}{3}\right)\pi \\ =100\sqrt{3}-50\pi \\ =173.20-157.07\\ =16.12\end{array}$

b.

To determine

To find: The shaded areawith radius of circle as 6.

Answer to Problem 36RP

The shaded area is $34.43$ .

Explanation of Solution

Given information:

Circle is of radius 6.

Formula used:

Area of equilateral triangle: $A=\frac{s^2}{4}\sqrt{3}$

s = side of equilateral triangle.

Area of sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$

r = radius of circle.

Area of triangle: $A=\frac{1}{2}\ast b\ast h$

b = base of triangle

h = height of triangle

Calculation:

  

Divide the shaded region into 3 parts and find the area of each part. The long triangle formed is a ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ .

Hypotenuse=12,

Side opposite to ${30}^{\circ }$ angle = 6,

Side opposite to ${60}^{\circ }$ angle = $6\sqrt{3}$

Region II is equiangular triangle, so all sides = 6.

Area of region I:

  $\begin{array}{l}{A}_{\mathrm{Ⅰ}}=A_{\mathrm{Sec}tor}-A_{EquilateralTriangle}\\ A_{\mathrm{Ⅰ}}=\left(\frac{measureofarc}{360^o}\right)\pi r^2-\frac{s^2}{4}\sqrt{3}\\ A_{\mathrm{Ⅰ}}=\left(\frac{{60}^{\circ }}{360^o}\right)\pi 6^2-\frac{6^2}{4}\sqrt{3}\\ A_{\mathrm{Ⅰ}}=\left(\frac{1}{6}\right)\pi 6^2-\frac{6^2}{4}\sqrt{3}\\ A_{\mathrm{Ⅰ}}=\left(\frac{1}{6}\right)36\pi -\frac{36}{4}\sqrt{3}\\ A_{\mathrm{Ⅰ}}=6\pi -9\sqrt{3}\end{array}$

Area of region II and III:

  $\begin{array}{l}{A}_{\mathrm{Ⅱ}+\mathrm{Ⅲ}}=A_{Triangle}\\ A_{\mathrm{Ⅱ}+\mathrm{Ⅲ}}=\frac{1}{2}\ast b\ast h\\ A_{\mathrm{Ⅱ}+\mathrm{Ⅲ}}=\frac{1}{2}\ast 6\ast 6\sqrt{3}\\ A_{\mathrm{Ⅱ}+\mathrm{Ⅲ}}=18\sqrt{3}\end{array}$

Shaded area = Area of region I + Area of region II and III

  $\begin{array}{l}{A}_{Shadedarea}=A_{\mathrm{Ⅰ}}+A_{\mathrm{Ⅱ}+\mathrm{Ⅲ}}\\ A_{Shadedarea}=\left(6\pi -9\sqrt{3}\right)+18\sqrt{3}\\ A_{Shadedarea}=6\pi +9\sqrt{3}\\ A_{Shadedarea}=34.43\end{array}$